Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 12}{x + 2} = \dfrac{-2x + 23}{x + 2}$
Answer: Multiply both sides by $x + 2$ $ \dfrac{x^2 - 12}{x + 2} (x + 2) = \dfrac{-2x + 23}{x + 2} (x + 2)$ $ x^2 - 12 = -2x + 23$ Subtract $-2x + 23$ from both sides: $ x^2 - 12 - (-2x + 23) = -2x + 23 - (-2x + 23)$ $ x^2 - 12 + 2x - 23 = 0$ $ x^2 - 35 + 2x = 0$ Factor the expression: $ (x - 5)(x + 7) = 0$ Therefore $x = 5$ or $x = -7$ The original expression is defined at $x = 5$ and $x = -7$, so there are no extraneous solutions.